What sort of tools are out there that can detect when a manifold is a product of other manifolds?
For example, comparing the homology of the circle to the torus, the homology of the torus gets more "complicated" as in the betti numbers for $H_1(T)$ is one higher than $H_1(S^1)$.
I was wondering if there are actual tools to detect products.
I can think of ways to detect that a space is not a product of other spaces, but getting a positive result seems extremely hard. In particular I don't know what I would do if a space were, say, proper homotopy equivalent to a product but not homeomorphic to a product.
First, by the Künneth theorem, the homology or cohomology of $X \times Y$ with coefficients in a field $k$ factors as the graded tensor product of the hoomology or cohomology of $X$ and $Y$ respectively. More concretely, if $p_X(t) = b_0 + b_1 t + b_2 t^2 + ...$ is the generating function of the Betti numbers $b_i = \dim H_i(X, k)$ of $X$ over $k$ (suppressing both $X$ and $k$ in the notation), then assuming all of the Betti numbers involved are finite we have
$$p_{X \times Y}(t) = p_X(t) p_Y(t)$$
so if a space $X$ has the property that $p_X(t)$ does not factor as the product of two polynomials with nonnegative integer coefficients (note that this is slightly weaker than being irreducible over $\mathbb{Z}$), then it cannot factor as the product of two other spaces in an "interesting" way (as long as we restrict to spaces with finitely generated homology / cohomology); in particular, if $X$ is a closed manifold, then it cannot factor as the product of two other closed manifolds of positive dimension. (In the noncompact case I think we can use Borel-Moore homology or equivalently compactly supported cohomology to get a similar result.)
Example. The orientable surface $\Sigma_g$ of genus $g$ has Poincare polynomial $1 + 2g t + t^2$, which is irreducible for $g \neq 1$. We conclude that $\Sigma_g$ cannot factor as a product of closed manifolds unless $g = 1$, where $\Sigma_1 \cong S^1 \times S^1$, corresponding to the factorization $1 + 2t + t^2 = (1 + t)^2$. (But of course we could also deduce this from the classification of closed $1$-manifolds.)
Unfortunately, as Ted Shifrin points out in the comments, often this will not be good enough to distinguish genuine products from fiber bundles. Roughly speaking, if a space $E$ is a fiber bundle over a space $B$ with fiber $F$, then $E$ is locally isomorphic to $B \times F$ but globally more interesting things can happen. The simplest example to visualize is the Möbius strip, which is a nontrivial line bundle over the circle $S^1$.
The Serre spectral sequence can be used to compute the homology and cohomology of fiber bundles, and in the nicest case it turns out that even if $E$ is not globally isomorphic to $B \times F$, it might still have the same homology. The Leray-Serre theorem describes one case where this happens (although it's a bit involved to state so I won't). In general it's very hard to tell when fiber bundles are trivial.
Example. The Lie group $\text{SU}(3)$ acts on $\mathbb{C}^3$ in a way preserving the unit sphere, which here is $S^5$. Moreover this action is transitive with stabilizer $\text{SU}(2)$, which is diffeomorphic to $S^3$. Hence we have a fiber bundle
$$S^3 \to \text{SU}(3) \to S^5$$
(that is, $\text{SU}(3)$ is a fiber bundle over $S^5$ with fiber $S^3$). Now it turns out that $\text{SU}(2)$ has the same homology and cohomology as the trivial bundle $S^3 \times S^5$ (even, I think, over $\mathbb{Z}$ and as a ring for cohomology). Nevertheless $\text{SU}(3)$ is not homeomorphic or even homotopic to $S^3 \times S^5$; see this MO question for details, where two arguments are given, one involving the computation of some $\pi_4$s and the other involving Steenrod operations.
Edit: Here's another obstruction: if a manifold factors as the product of other manifolds, then its tangent bundle factors as a direct sum of other bundles. There are obstructions to this being possible coming from characteristic classes. For example, you can prove that the tangent bundle of $S^2$ is not a direct sum of any other bundles, and this gives another proof that $S^2$ is not a product.